# Math Help - Solving a Matrix

1. ## Solving a Matrix

Find all values $a, b, c$ given the equations below such that there (a) exists exactly 1 solution (b) exists an infinite # of solutions (c) exists no solution

$x_1 + 5x_2 + x_3 = 0$
$x_1 + 6x_2 - x_3 = 0$
$2x_1 + ax_2 + bx_3 = c$

My work for this problem:

We have \left[ \begin {array}{cccc} 1&5&1&0\\\noalign{\medskip}1&6&-1&0\\\noalign{\medskip}2&a&b&c\end {array} \right]

Using Maple to get it in RREF we have:

\left[ \begin {array}{cccc} 1&0&0&{\frac {-11c}{b-22+2\,a}}\\\noalign{\medskip}0&1&0&{\frac {2c}{b-22+2\,a}}\\\noalign{\medskip}0&0&1&{\frac {c}{b-22+2\,a}}\end {array} \right]

It looks like it will always have a solution (since there is a pivot in every row). And the only way to get an infinite # of solutions would be to have a free variable. So I'm not sure what to do with this problem.

2. I noticed for this one its not helpful getting it in RREF but simpy just echelon form (gausselim), which is:

\left[ \begin {array}{cccc} 1&5&1&0\\\noalign{\medskip}0&1&-2&0\\\noalign{\medskip}0&0&b-22+2\,a&c\end {array} \right]

I think that should help.

Based on the latex representation from Maple, I think the a might be in the augmented column but its not clear (doesn't look like it though).

3. Meh I thought this would be an easy linear algebra question. I guess not. I know that it's inconsistent if c = 1 and if b - 22 + 2a = 0.

Getting exactly 1 solution would mean b - 22 + 2a = 1 I believe.

And getting an infinite # of solutions is impossible I think b/c we'll never have free variables.

4. Originally Posted by LinAlg
I noticed for this one its not helpful getting it in RREF but simpy just echelon form (gausselim), which is:

\left[ \begin {array}{cccc} 1&5&1&0\\\noalign{\medskip}0&1&-2&0\\\noalign{\medskip}0&0&b-22+2\,a&c\end {array} \right]

I think that should help.

Based on the latex representation from Maple, I think the a might be in the augmented column but its not clear (doesn't look like it though).
take it a step further. remember we have infinitely many solutions if one row is a multiple of the other. so get the last row as a multiple of the second. i assume your matrix is right up to this point, so i will just continue from there. take the second row minus the third and write that as the new third row, we get

$\left[ \begin{array}{cccc} 1 & 5 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 1 & 20 - b - 2a & -c \end{array} \right]$

now, if $c = 0$ and at the same time $20 - b - 2a = -2$, then the second row wipes out the third and we have no pivot for the third, yielding an infinite number of solutions.

as you said earlier, going by your matrix, $b - 22 + 2a = 1$ and $c = \text{whatever}$ yields a single solution

while $b - 22 + 2 = 0$ and $c \ne 0$ yields an inconsistent system, and hence we have no solutions